Question

Giải pt:

\(x^3+3x+2=\left(x+2\right)\sqrt{x^3+2x+1}\)

Answer 1

Lời giải:

Đặt \(\sqrt{x^3+2x+1}=a\Rightarrow x^3+2x+1=a^2\Rightarrow x^2+3x+2=a^2+x+1\)

PT trở thành:

\(a^2+x+1=(x+2)a\)

\(\Leftrightarrow (a^2-2a+1)+(x-xa)=0\)

\(\Leftrightarrow (a-1)^2-x(a-1)=0\)

\(\Leftrightarrow (a-1)(a-1-x)=0\) \(\Rightarrow \left[\begin{matrix} a=1\\ a=x+1\end{matrix}\right.\)

Nếu \(a=1\Rightarrow x^3+2x=0\Leftrightarrow x(x^2+2)=0\Rightarrow x=0\)

Nếu \(a=x+1\Rightarrow x^3+2x+1=a^2=(x+1)^2=x^2+2x+1\)

\(\Rightarrow x^3=x^2\Leftrightarrow x^2(x-1)=0\Rightarrow \left[\begin{matrix} x=0\\ x=1\end{matrix}\right.\)

Vậy.......